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### Messages - Qi Zeng

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##### Final Exam / Re: FE-P1
« on: December 18, 2018, 07:16:16 AM »
\begin{align*}
f(z) &= \frac{1}{(z-(-1+i))(z-(-1-i))}\\
&= -\frac{1}{2}i\frac{1}{z-(-1+i)} + \frac{1}{2}i\frac{1}{z-(-1-i)}\\
\end{align*}

(a)
\begin{align*}
f(z) &= -\frac{i}{2(-1+i)}\times\frac{1}{\frac{z}{(-1+i)}-1} + \frac{i}{2(-1-i)}\times\frac{1}{\frac{z}{(-1-i)}-1}\\
&=\frac{i}{2(-1+i)}\times \sum\limits_{n=0}^{\infty}(\frac{z}{-1+i})^n - \frac{i}{2(-1-i)}\times\sum\limits_{n=0}^{\infty}(\frac{z}{-1-i})^n
\end{align*}

The function converges when $|\frac{z}{-1+i}| < 1$ and $|\frac{z}{-1-i}| < 1$, therefore $\frac{|z|}{\sqrt{2}} < 1$, and $|z| < \sqrt{2}$.

The radius of convergence r = $\sqrt{2}$, and it diverges when $|z|$ = $\sqrt{2}$.

(b)
\begin{align*}
f(z) &= -\frac{i}{2z}\times\frac{1}{1 - \frac{-1+i}{z}} + \frac{i}{2z}\times\frac{1}{1 - \frac{-1-i}{z}}\\
&=-\frac{i}{2z}\times \sum\limits_{n=0}^{\infty}(\frac{-1+i}{z})^n + \frac{i}{2z}\times\sum\limits_{n=0}^{\infty}(\frac{-1-i}{z})^n
\end{align*}

The function converges when $|\frac{-1+i}{z}| < 1$ and $|\frac{-1-i}{z}| < 1$, therefore $\frac{\sqrt{2}}{|z|} < 1$, and $|z| > \sqrt{2}$.

The radius R = $\sqrt{2}$, and it diverges when $|z|$ = $\sqrt{2}$.

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